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3 years ago

5

A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring

the car to a halt in a distance of 33.1 m?

1 answer:

AleksandrR [38]3 years ago

5 0

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

$v^2-u^2=2as$

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

$0-(17.7)^2=2a\times 33.1$

$a=\dfrac{-(17.7)^2}{33.1}$

$a=-9.46\ m/s^2$

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

$F = ma$

$F =1400\times(-9.46)$

$F=-13244\ N$

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

$|F| =13244\ N$

Hence, The magnitude of the horizontal net force is 13244 N.

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CHECK COMPLETE QUESTION BELOW

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C) the same as you do on Earth

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Explanation :

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Read 2 more answers

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

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